Calculate the energy released when 8000 small water droplets, each of same radius 2×10−7m coalesce to form one large drop. The surface tension of water is 7.5×10−2N/m
A
4.2×10−10J
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B
2.8×10−10J
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C
zero
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D
6.5×10−10J
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Solution
The correct option is B2.8×10−10J Let r be the radius of smaller drops and R of the bigger one.
Equating the initial and final volume, we have ⇒43πR3=8000×(43πr3) ⇒R3=8000r3 ⇒R=(8000)13r ⇒R=20r=20×2×10−7=4×10−6m
Increase in surface area ΔA=4πR2−8000(4πr2) =4π×400r2−8000×4πr2 =−7600×4π×r2 =−7600×4π×4×10−14
[-ve sign shows surface area is decreasing ]
Energy released in the process U=T|ΔA|=7.5×10−2×7600×16π×10−14 =2.8×10−10J